Abstractly, on the topological circle $S^1$ there are only two real line bundles, up to isomorphism: the trivial one $\mathcal{O}$ and the Moebius strip $\mathcal{O}(1)$ (thinking of $S^1$ as $\mathbb{RP}^1$). So we have $\mathcal{O}(2k)\cong\mathcal{O}$ and $\mathcal{O}(2k+1)\cong\mathcal{O}(1)$ for any $k$.

Let's embed $S^1\hookrightarrow\mathbb{R}^3$ as the set $x^2+y^2=1$ in the $z=0$ plane. We can embed the total spaces $\iota_n:\mathrm{tot}(\mathcal{O}(n))\hookrightarrow\mathbb{R}^3$, respecting the zero section, for example as the union, for $t\in [0,2\pi]$, of the (open) segments joining the two points

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No. The unit normal bundle to the circle is a torus. Each fiber of the Mobius strip contains a unit vector going in some choice of direction, lying on that torus. That vector travels around the torus representing a homology class, which is clearly not trivial. If we can isotope to the trivial strip, we get a homotopy to the trivial homology class.

You could look at the knot type of the boundary of these twisted bands. (Substitute the unit sphere bundle if you want the noncompact version.) For each of your $\mu(k)$ you get a $T(2,k)$ torus link as the boundary. Each of these is non-isotopic. For example, $k=2$ gives the Hopf link and $k=3$ the trefoil. This implies that no two would be ambiently isotopic.